Brain hurts, help with math

[coxcomb]

Hi All,
I have seen some math whiz types posting here before and I hope one of them can help me.

The basic idea is that you roll a number of d12s looking for results that are less than or equal to a target number. So, say, I'm rolling 4d12 and each die that comes up 6 or less is a success.

What I need to do is determine the odds of getting each possible number of successes for each rolling combination up to rolling 12d12 looking for 12 or less (chance for 12 successes = 100% -- see, I'm not THAT hopeless).

Any help?

[Jack Aidley]

Let's let s be your chance of success (target/12) and f be your chance of failure (1-s). Then your probability of getting n successes on d dice is:

(s^n)*(f^(d-n))*number of possible arrangements

I'm using ^ to mean to the power of... (why does BBCode not have super- and sub- script?)

So how many possible arrangement are there? Well for a set of n distinct objects, the number of possible arrangements is n!, but when some of the items are identical, this is divided by the factorial of that number, so in your case:

possible arrangements = d!/(n!(d-n)!) (which happens to equal dCn)

So, for 3 dice, with a TN of 5:

0 successes = (7/12)^3*1 / 12^3 ~= 0.20
1 success = (5/12)*(7/12)^2*3 ~= 0.43
2 successes = (5/12)^2*(7/12)*3 ~= 0.30
3 successes = (5/12)^3 ~= 0.07

That help?

[coxcomb]

Thanks, Jack, you're my new hero!